# How to find Rotational and Translational Equilibrium of Hanging Masses on a Bar?

I am making a hanging mobile which needs to be done mathematically by calculating torque. The problem is, I can't seem to figure out how to solve for the distance of the two masses from the pivot point (which is a mass itself).

Here is what I was given to work on this:

1. Forceup = Forcedown
2. Torqueclockwise = Torquecounterclockwise
3. Use an object as the pivot point if there is more than one object to solve for.

Here is what I filled in those steps:

1. Forceup = (2.3814 kg + .0452 kg + .0878 kg + 0.0284 kg)9.8 m/s2 = 3.96312 N.
2. 2.3814 kg(9.8 m/s2)(0 m.) + 0.0284 kg(9.8 m/s2)(0.125 m.) + 0.0878 kg(9.8 m/s2)(x m.) + 0.0452 kg(9.8 m/s2)(y m.) = 3.96312 N.(.125 m.)

which, if my math is correct, is:

0.03479 Nm. + 0.86044 N (x m.) + 0.44296 N (y m.) = 0.49539 Nm.

Note: 0.0284 kg is the weight of the rod that is holding up the objects and the string holding the rod up is in the center of the rod, which is why the force up is the same distance away from the pivot point as the rod.

So my problem lies here: what do I do with the two different variables? I don't have a second equation nor do I know of a second equation that I can use to calculate the distance needed between them. What can I do to make this work?

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 It's a little hard to tell what the setup is. A bar hanging from a single string with 3 masses on the bar? Or are there two bars? Could you clarify a little? It's entirely possible to have situations without a unique solution: pick an $x$ to locate one object, and solve for the $y$ to place the other to keep everything in balance. But I can't be sure if that's the case here. – Chris White Jan 7 at 5:12